我有一个大串像
res = ["FAV_VENUE_CITY_NAME == 'Mumbai' & EVENT_GENRE == 'KIDS' & count_EVENT_GENRE >= 1",
"FAV_VENUE_CITY_NAME == 'Mumbai' & EVENT_GENRE == 'FANTASY' & count_EVENT_GENRE >= 1",
"FAV_VENUE_CITY_NAME =='Mumbai' & EVENT_GENRE == 'FESTIVAL' & count_EVENT_GENRE >= 1",
"FAV_VENUE_CITY_NAME == 'New Delhi' & EVENT_GENRE == 'WORKSHOP' & count_EVENT_GENRE >= 1",
"FAV_VENUE_CITY_NAME == 'Mumbai' & EVENT_GENRE == 'EXHIBITION' & count_EVENT_GENRE >= 1",
"FAV_VENUE_CITY_NAME == 'Bangalore' & FAV_GENRE == '|DRAMA|'",
"FAV_VENUE_CITY_NAME = 'Mumbai' & & FAV_GENRE == '|ACTION|ADVENTURE|SCI-FI|'",
"FAV_VENUE_CITY_NAME == 'Bangalore' & FAV_GENRE == '|COMEDY|'",
"FAV_VENUE_CITY_NAME == 'Bangalore' & FAV_GENRE == 'DRAMA' & FAV_LANGUAGE == 'English'",
"FAV_VENUE_CITY_NAME == 'New Delhi' & FAV_LANGUAGE == 'Hindi' & count_EVENT_LANGUAGE >= 1"]
现在我正在提取字段
res = [re.split(r'[(==)(>=)]', x)[0].strip() for x in re.split('[&($#$)]', whereFields)]
res = [x for x in list(set(res)) if x]
o/p:['FAV_GENRE', 'FAV_LANGUAGE', 'FAV_VENUE_CITY_NAME', 'count_EVENT_GENRE', 'EVENT_GENRE','count_EVENT_LANGUAGE']
我正在获得价值
FAV_VENUE_CITY_NAME = ['New Delhi', 'Mumbai', 'Bangalore']
FAV_GENRE = ['|DRAMA|', '|COMEDY|', '|ACTION|ADVENTURE|SCI-FI|', 'DRAMA']
EVENT_GENRE = ['FESTIVAL', 'WORKSHOP', 'FANTASY', 'KIDS', 'EXHIBITION']
FAV_LANGUAGE = ['English', 'Hindi']
count_on_field = ['EVENT_GENRE', 'EVENT_LANGUAGE']
现在，我想制作一本字典，其关键字将是res中的字段名称。值将是以上链接的结果。
或者有没有办法使列表项本身成为不同的不同列表。
就像是
res = ['FAV_GENRE', 'FAV_LANGUAGE', 'FAV_VENUE_CITY_NAME', 'count_EVENT_GENRE', 'EVENT_GENRE','count_EVENT_LANGUAGE']
for i in range(len(res)):
res[i] = list(res[i]) # make each item as an empty list with name as it is
这样他们就变得像
FAV_VENUE_CITY_NAME = []
EVENT_GENRE = []
FAV_GENRE = []
FAV_LANGUAGE = [
然后按照上述链接中的方法将值获取到res列表中的每个单独列表。
然后制作一个像下面这样的字典，以索引为键的字典
a = [51,27,13,56]
b = dict(enumerate(a))
#####d = dict{key=each list name from res list, value = value in each ind. lists}＃
或如果可能的话，建议使用顶级资源列表中的内容。...如何形成字典，将键作为字段名称，将值作为每行的值
o/p: d = {'FAV_VENUE_CITY_NAME':['Mumbai','New Delhi','Bangalore'], 'EVENT_GENRE':['KIDS','FANTASY','FESTIVAL','WORKSHOP','EXHIBITION'], 'FAV_GENRE':['|DRAMA|','|ACTION|ADVENTURE|SCI-FI|','|COMEDY|','DRAMA'], 'FAV_LANGUAGE':['English','Hindi']}
count_EVENT_GENRE> = 1，count_EVENT_LANGUAGE> = 1不应在该词典中，而应转到列表中
count_on_fields = ['EVENT_GENRE','EVENT_LANGUAGE']
如果有人有更好的主意或建议，请帮助。
解决方案
干得好：
创建具有所有值的列表：
values=[
FAV_GENRE,
FAV_LANGUAGE,
FAV_VENUE_CITY_NAME,
EVENT_GENRE,
count_on_field
]
然后根据此答案的建议创建您的字典：
d=dict(zip(res, values))
请注意，数组顺序确实很重要...
尚未测试，因为我的电池快没电了。我希望它能满足您的需求
python创建一个字典 其中值为1_从2个列表创建一个字典 其中一个作为键 另一个作为python中的值...
